MomentumOperators and ObservablesRecall the time-independent Schrodinger equation:

E is the total energy of the particle:

E = KE +U =p2

2m+U

!!2

2m

"2

"x2+U(x)

#

$%

&

'()(x) = E)(x)

Comparing the two equations we see that:

!!2

2m

"2

"x2= KE =

p2

2m# p = !i!

"

"x

Momentum is described by a derivative operation in theposition representation.

Momentum

Every observable quantity is described by an operator which canoperate on wave functions.

For position wave functions ψ(x),

For momentum wave functions ζ(p),

Observables that are functions of x and p are described by operators thatare corresponding functions of the x and p operators

Position : x (operation :multiply by x)

Momentum : p = !i!" "x (operation :partial derivative)

Position : i!! !p

Momentum : p

Operators and Observables

MomentumObservables and corresponding operators (position wave functions)

MomentumExpectation values

What about the mean value of p ? Recall that

f (x) = f (x)P(x)dx =!"

"

# f (x)$ (x,t)2dx

!"

"

#

p = mv = mdx

dt! p = m

d x

dt

x = xP(x)dx =

!"

"

# x $ (x,t)2dx

!"

"

#

Using

And taking the time derivative, after some manipulation usingthe Schrodinger equation we get,

p = ! *!

i

"!"x

dx#$

$

% = ! *p!dx

#$

$

%

MomentumExpectation values

In general the average value of any observable O is

Where is the associated operator.

p = ! *!

i

"!"x

dx#$

$

% = ! *p!dx

#$

$

%

O == ! *O!dx

"#

#

$

O

Momentum DistributionAny function can be written as a sum of cosine and sine waves:

Periodic function: Fourier Series:Spatial function at a fixed time:

f (x) = ane

iknx

n=!"

"

# = an

cos(knx) + isin(k

nx)( )

n=!"

"

#

an=

1

2Lf (x)e

! iknxdx

! L

L

" =1

2Lf (x) cos(k

nx) ! isin(k

nx)( )dx

! L

L

"

V (t) = ane

ik!nt

n="#

#

$ = an

cos(!nt) + isin(!

nt)( )

n="#

#

$

an=

1

2TV (t)e

! i"ntdt

!T

T

# =1

2TV (t) cos("

nt) ! isin("

nt)( )dt

!T

T

#

Temporal function at a fixed position:

Non-periodic function: Fourier Integral

f (x) =1

2!a(k)eikxdk

"#

#

$ =1

2!a(k) cos(kx) + isin(kx)( )dk

"#

#

$

a(k) =1

2!f (x)e" ikxdx

"#

#

$ =1

2!f (x) cos(kx) " isin(kx)( )dx

"#

#

$

V (t) =1

2!g(" )ei" td"

#$

$

% =1

2!g(" ) cos("t) + isin("t)( )d"

#$

$

%

g(! ) =1

2"V (t)e# i! tdt

#$

$

% =1

2"V (t) cos(!t) # isin(!t)( )dt

#$

$

%

Spatial function at a fixed time:

Temporal function at a fixed position:

Momentum Distribution

See http://www.falstad.com/fourier/ for a demonstration of Fourierseries

Examples:

Momentum Distribution

Fourier decomposition of some functions:

f(x) a(k)

Acos k1x( ) + Acos k2x( )

f(x) a(k)

Momentum Distribution

Gaussian Gaussian

f x( ) = C exp!(x ! xo )2

2" x2

#

$%%

&

'((

Fourier decomposition of some functions:

f(x) a(k)

a k( ) = N exp!(k ! k

o)2

2"k

2

#

$%%

&

'((

x

k

!x!k " 1, !#!t " 1

Recall the reciprocity relations:

For Gaussian functions, the uncertainties are minimized

!x!k =

1

2, !"!t =

1

2

Momentum Distribution

Heisenberg Uncertainty Principle

Gaussian Gaussian

p = !k, E = !!

!p!x "

!

2

x

k

Minimum uncertainty:

• Our knowledge of conjugate quantities is inherently uncertain.• No matter how good our measuring instruments are we cannot

simultaneously know x and p or E and t with complete precision.

!x!k =

1

2, !"!t =

1

2

!E!t "!

2

For Gaussian functions:

!p!x =

!

2

!E!t =!

2

Momentum space wave functionsSince p=hk/2π, the functions a(k) can be easily written as functions ζ(p).Given a wave function in coordinate space representing the position of aparticle, the corresponding momentum wave function is the Fourierintegral of the the position wave function:

If the position wave function is normalized, the momentum wavefunction is also normalized to unity:

! (p,t) =1

2"!( )1/2

e

# i!

xp( )$ (x,t)% dx

! (p,t)"2

dp = 1

Momentum Distribution

Momentum space wave functions

We can interpret the square of the quantity

of the normalized wave function as the probability at time t of themomentum of a particle being in the volume dp around p

P(p,t) = ! (p,t)2dp

Hence the normalization requirement is a statement that theprobability of finding the particle with some value of momentum is 1.

! (p,t)"2

dx = P(p,t)dp =" 1

Momentum Distribution

Momentum space wave functions• The probability of finding a particle with momentum a < p < b isthen

! (p,t)2

a

b

" dp

This is the area under the momentum probability density curvebetween p=a and p=b:

p

!(p,t)2

Momentum Distribution

Given a wave function representing the momentum of a particle, thecorresponding position space wave function is the inverse Fourierintegral of the momentum wave function:

If the momentum wave function is normalized, the position wavefunction is also normalized to unity.

Note that the position and momentum wave functions represent thesame state of the system. All information about the state can beobtained from either wave function.

! (x,t) =1

2"!( )1/2

e

i

!px( )

# $ (p,t)dp

Momentum Distribution